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Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction 
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#1
Sep1411, 05:29 PM

P: 35

1. The problem statement, all variables and given/known data
Two horses pull horizontally on ropes attached to a tree stump. Each horse pulls with a force of magnitude F. If the resultant force (vector R) has the magnitude R = 1.51 F, what is the angle (in degrees) between the two ropes? 2. Relevant equations 3. The attempt at a solution Ry= 0+cos Rx= 1+ sin R= 1.51 sin=cos/1.54 tan= 1/1.54 angle= 33 degrees 33 * 2 = total angle for both of 66 degrees 


#2
Sep1411, 05:42 PM

P: 35

I'm on my 5th hour for this problem and still can't get it. I realize it's necessary to assign an arbitrary value of say 1 to F. That gives F=1 and R=1.51, but everything I've tried to do with that information has failed.



#3
Sep1411, 06:02 PM

P: 58

Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G.
So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1). F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees That's just half of your angle though since you split it into two triangles 49.025*2 = 98.05 degrees 


#4
Sep1411, 06:15 PM

P: 35

Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction



#5
Sep1411, 06:25 PM

P: 58

Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.



#6
Sep1411, 06:32 PM

P: 35




#7
Sep1411, 07:35 PM

P: 35




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